Samacheer Kalvi 10th Maths Exercise 2.3 Solutions

10th Maths Exercise 2.3 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 14 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance13 answers · full steps below

Q1 (i) $x = 7$
Q1 (ii) $x = 5$
Q1 (iii) $x = 2$
Q1 (iv) $x = 7$
Q1 (v) $x = 2$
Q2 $7x - 3 \equiv 3 \pmod{17}$
Q3 $x = 2, 8, 14, 20, \ldots$ i.e. $x \equiv 2 \pmod 6$
Q4 $x = 8, 19, 30, 41, \ldots$ i.e. $x \equiv 8 \pmod{11}$
Q511 a.m.
Q68 a.m.
Q7Friday
Q9Remainder $= 2$
Q10The flight lands at London on Monday at 6:00 a.m. (London time)

Question 1

(i)

Find the least positive value of $x$ such that $71 \equiv x \pmod 8$

$71 \equiv x \pmod 8$ means $71 - x$ is a multiple of $8$ .
Divide $71$ by $8$ : $71 = 8 \times 8 + 7$ .
So the remainder is $7$ .

Answer

$x = 7$

(ii)

Find the least positive value of $x$ such that $78 + x \equiv 3 \pmod 5$

$78 + x \equiv 3 \pmod 5$ means $78 + x - 3 = 5n$ .
$75 + x = 5n$ , so $75 + x$ must be a multiple of $5$ .
$75$ itself is a multiple of $5$ , so the next multiple is $80$ .
$\therefore x = 80 - 75 = 5$ .

Answer

$x = 5$

(iii)

Find the least positive value of $x$ such that $89 \equiv (x + 3) \pmod 4$

$89 \equiv (x+3) \pmod 4$ means $89 - (x+3) = 4n$ .
$86 - x = 4n$ , so $86 - x$ is a multiple of $4$ .
The nearest multiple of $4$ less than $86$ is $84$ .
$\therefore x = 86 - 84 = 2$ .

Answer

$x = 2$

(iv)

Find the least positive value of $x$ such that $96 \equiv \dfrac{x}{7} \pmod 5$

$96 \equiv \dfrac{x}{7} \pmod 5$ means $96 - \dfrac{x}{7} = 5n$ .
Multiply by $7$ : $672 - x = 35n$ , so $672 - x$ is a multiple of $35$ .
The nearest multiple of $35$ less than $672$ is $35 \times 19 = 665$ .
$\therefore x = 672 - 665 = 7$ .

Answer

$x = 7$

(v)

Find the least positive value of $x$ such that $5x \equiv 4 \pmod 6$

$5x \equiv 4 \pmod 6$ means $5x - 4 = 6k$ .
$x = \dfrac{6k + 4}{5}$ .
Put $k = 1$ : $x = \dfrac{10}{5} = 2$ .
So the least positive value is $2$ .

Answer

$x = 2$

Q2. If $x$ is congruent to $13$ modulo $17$ then $7x - 3$ is congruent …

If $x$ is congruent to $13$ modulo $17$ then $7x - 3$ is congruent to which number modulo $17$ ?

$x \equiv 13 \pmod{17}$ .
Multiply by $7$ : $7x \equiv 91 \pmod{17}$ .
Subtract $3$ : $7x - 3 \equiv 88 \pmod{17}$ .
$88 = 17 \times 5 + 3$ , so $88 \equiv 3 \pmod{17}$ .
$\therefore 7x - 3 \equiv 3 \pmod{17}$ .

Answer

$7x - 3 \equiv 3 \pmod{17}$

Q3. Solve $5x \equiv 4 \pmod 6$

$5x \equiv 4 \pmod 6$ can be written as $5x - 4 = 6k$ for some integer $k$ .
$x = \dfrac{6k + 4}{5}$ .
$k = 1 \Rightarrow x = 2$ ; $k = 6 \Rightarrow x = 8$ ; $k = 11 \Rightarrow x = 14$ .
The solutions are $2, 8, 14, 20, \ldots$ , that is $x \equiv 2 \pmod 6$ .

Answer

$x = 2, 8, 14, 20, \ldots$ i.e. $x \equiv 2 \pmod 6$

Q4. Solve $3x - 2 \equiv 0 \pmod{11}$

$3x - 2 \equiv 0 \pmod{11}$ gives $3x \equiv 2 \pmod{11}$ .
So $3x - 2 = 11k$ , that is $x = \dfrac{11k + 2}{3}$ .
$k = 2 \Rightarrow x = 8$ ; $k = 5 \Rightarrow x = 19$ ; $k = 8 \Rightarrow x = 30$ .
The solutions are $8, 19, 30, 41, \ldots$ , that is $x \equiv 8 \pmod{11}$ .

Answer

$x = 8, 19, 30, 41, \ldots$ i.e. $x \equiv 8 \pmod{11}$

Q5. What is the time $100$ hours after 7 a.m.?

A day has $24$ hours, so work with modulo $24$ .
$7 + 100 = 107$ .
$107 = 24 \times 4 + 11$ , so $107 \equiv 11 \pmod{24}$ .
The time is $11$ hours, i.e. 11 a.m. (after $4$ days).

Answer

11 a.m.

Q6. What is the time $15$ hours before 11 p.m.?

11 p.m. is $23$ hours in $24$ -hour form.
$23 - 15 = 8$ .
$8 \equiv 8 \pmod{24}$ .
The time is $8$ hours, i.e. 8 a.m.

Answer

8 a.m.

Q7. Today is Tuesday. My uncle will come after $45$ days. In which …

Today is Tuesday. My uncle will come after $45$ days. In which day my uncle will be coming?

Give the numbers $0,1,2,3,4,5,6$ to Sunday, Monday, …, Saturday.
Today is Tuesday $= 2$ . Work with modulo $7$ (a week has $7$ days).
$2 + 45 = 47$ .
$47 = 7 \times 6 + 5$ , so $47 \equiv 5 \pmod 7$ .
The number $5$ stands for Friday.

Answer

Friday

Q8. Prove that $2^{n} + 6 \times 9^{n}$ is always divisible by $7$ for any positive integer …

Prove that $2^{n} + 6 \times 9^{n}$ is always divisible by $7$ for any positive integer $n$ .

Show steps (6)

$9 - 2 = 7$ is divisible by $7$ , so $9 \equiv 2 \pmod 7$ .
Raising to the power $n$ : $9^{n} \equiv 2^{n} \pmod 7$ .
Multiply by $6$ : $6 \times 9^{n} \equiv 6 \times 2^{n} \pmod 7$ .
Add $2^{n}$ to both sides: $2^{n} + 6 \times 9^{n} \equiv 2^{n} + 6 \times 2^{n} \pmod 7$ .
$2^{n} + 6 \times 2^{n} = 7 \times 2^{n} \equiv 0 \pmod 7$ .
$\therefore 2^{n} + 6 \times 9^{n} \equiv 0 \pmod 7$ for every positive integer $n$ .

Answer

$2^{n} + 6 \times 9^{n}$ is divisible by $7$ for any positive integer $n$ .

Q9. Find the remainder when $2^{81}$ is divided by $17$ .

$2^{4} = 16 \equiv -1 \pmod{17}$ .
So $2^{8} = (2^{4})^{2} \equiv (-1)^{2} = 1 \pmod{17}$ .
$81 = 8 \times 10 + 1$ , so $2^{81} = (2^{8})^{10} \times 2$ .
$2^{81} \equiv 1^{10} \times 2 = 2 \pmod{17}$ .

Answer

Remainder $= 2$

Q10. The duration of flight travel from Chennai to London through British Airlines …

The duration of flight travel from Chennai to London through British Airlines is approximately $11$ hours. The airplane begins its journey on Sunday at $23{:}30$ hours. If the time at Chennai is four and half hours ahead to that of London's time, then find the time at London, when will the flight lands at London Airport.

Departure at Chennai time $23{:}30$ on Sunday, flight time $11$ hours. Use modulo $24$ .
Arrival in Chennai time $= 23{:}30 + 11 = 34{:}30 \equiv 10{:}30 \pmod{24}$ , i.e. Monday $10{:}30$ hours.
London time is $4$ hours $30$ minutes behind Chennai time.
$10{:}30 - 4{:}30 = 6{:}00$ hours.

Answer

The flight lands at London on Monday at 6:00 a.m. (London time)

Quick check

If $A = 2^{65}$ and $B = 2^{64} + 2^{63} + 2^{62} + \cdots + 2^{0}$ which of the following is true?
1. (1) $B$ is $2^{64}$ more than $A$
2. (2) $A$ and $B$ are equal
3. (3) $B$ is larger than $A$ by $1$
4. (4) $A$ is larger than $B$ by $1$
Show answer
Answer: (4) $A$ is larger than $B$ by $1$
1. $B = 2^0 + 2^1 + \cdots + 2^{64}$ is a G.P. with $a = 1$ , $r = 2$ , $n = 65$ .
2. $B = \dfrac{1(2^{65} - 1)}{2 - 1} = 2^{65} - 1$ .
3. $A - B = 2^{65} - (2^{65} - 1) = 1$ .
Question 1 of 5
The sum of the exponents of the prime factors in the prime factorization of $1729$ is
1. (1) $1$
2. (2) $2$
3. (3) $3$
4. (4) $4$
Show answer
Answer: (3) $3$
1. $1729 = 7 \times 13 \times 19 = 7^1 \times 13^1 \times 19^1$ .
2. Sum of exponents $= 1 + 1 + 1 = 3$ .
Question 2 of 5
If $6$ times of $6^{\text{th}}$ term of an A.P. is equal to $7$ times the $7^{\text{th}}$ term, then the $13^{\text{th}}$ term of the A.P. is
1. (1) $0$
2. (2) $6$
3. (3) $7$
4. (4) $13$
Show answer
Answer: (1) $0$
1. $6t_6 = 7t_7 \Rightarrow 6(a+5d) = 7(a+6d)$ .
2. $6a + 30d = 7a + 42d \Rightarrow a + 12d = 0$ .
3. $t_{13} = a + 12d = 0$ .
Question 3 of 5
An A.P. consists of $31$ terms. If its $16^{\text{th}}$ term is $m$ , then the sum of all the terms of this A.P. is
1. (1) $16\,m$
2. (2) $62\,m$
3. (3) $31\,m$
4. (4) $\dfrac{31}{2}\,m$
Show answer
Answer: (3) $31\,m$
1. $t_{16} = a + 15d = m$ .
2. $S_{31} = \dfrac{31}{2}\,[2a + 30d] = \dfrac{31}{2} \times 2(a + 15d)$ .
3. $= 31(a+15d) = 31m$ .
Question 4 of 5
Using Euclid's division lemma, if the cube of any positive integer is divided by $9$ then the possible remainders are
1. (1) $0, 1, 8$
2. (2) $1, 4, 8$
3. (3) $0, 1, 3$
4. (4) $1, 3, 5$
Show answer
Answer: (1) $0, 1, 8$
1. Any positive integer is of the form $3q$ , $3q+1$ or $3q+2$ .
2. $(3q)^3 = 27q^3 = 9(3q^3) \Rightarrow$ remainder $0$ .
3. $(3q+1)^3 = 27q^3+27q^2+9q+1 = 9(3q^3+3q^2+q)+1 \Rightarrow$ remainder $1$ .
4. $(3q+2)^3 = 27q^3+54q^2+36q+8 = 9(3q^3+6q^2+4q)+8 \Rightarrow$ remainder $8$ .
Question 5 of 5

Chapter 2 formulas

Euclid's division lemma
$a = bq + r,\ 0 \le r < |b|$
Fundamental theorem of arithmetic
$N = p_1^{a_1} p_2^{a_2} p_3^{a_3} \cdots p_k^{a_k}\ \text{(unique except for order)}$
Arithmetic Progression
$a,\ a+d,\ a+2d,\ a+3d,\ \ldots$
n-th term of an A.P.
$t_n = a + (n-1)d$
Sum to first n terms of an A.P.
$S_n = \frac{n}{2}\left[2a + (n-1)d\right]$
Sum of an A.P. when the last term l is given
$S_n = \frac{n}{2}\left[a + l\right]$
Geometric Progression
$a,\ ar,\ ar^2,\ \ldots,\ ar^{n-1}$
n-th term of a G.P.
$t_n = ar^{n-1}$
Sum to first n terms of a G.P. (r ≠ 1)
$S_n = \frac{a(r^n - 1)}{r - 1}$
Sum to first n terms of a G.P. when r = 1
$S_n = na$
Sum to infinite terms of a G.P. (−1 < r < 1)
$S_\infty = \frac{a}{1 - r}$
Sum of first n natural numbers
$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$
Sum of squares of first n natural numbers
$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$
Sum of cubes of first n natural numbers
$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$
Sum of first n odd natural numbers
$1 + 3 + 5 + \cdots + (2n-1) = n^2$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

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Question 1

(i)

(ii)

(iii)

(iv)

(v)

Q2. If xxx is congruent to 131313 modulo 171717 then 7x−37x - 37x−3 is congruent …

Q3. Solve 5x≡4(mod6)5x \equiv 4 \pmod 65x≡4(mod6)

Q4. Solve 3x−2≡0(mod11)3x - 2 \equiv 0 \pmod{11}3x−2≡0(mod11)

Q5. What is the time 100100100 hours after 7 a.m.?

Q6. What is the time 151515 hours before 11 p.m.?

Q7. Today is Tuesday. My uncle will come after 454545 days. In which …

Q8. Prove that 2n+6×9n2^{n} + 6 \times 9^{n}2n+6×9n is always divisible by 777 for any positive integer …

Q9. Find the remainder when 2812^{81}281 is divided by 171717 .

Q10. The duration of flight travel from Chennai to London through British Airlines …

Quick check

Q2. If $x$ is congruent to $13$ modulo $17$ then $7x - 3$ is congruent …

Q3. Solve $5x \equiv 4 \pmod 6$

Q4. Solve $3x - 2 \equiv 0 \pmod{11}$

Q5. What is the time $100$ hours after 7 a.m.?

Q6. What is the time $15$ hours before 11 p.m.?

Q7. Today is Tuesday. My uncle will come after $45$ days. In which …

Q8. Prove that $2^{n} + 6 \times 9^{n}$ is always divisible by $7$ for any positive integer …

Q9. Find the remainder when $2^{81}$ is divided by $17$ .